An ultrasonic fatigue test is a fatigue testing method which uses resonance with free vibration of a test specimen (NPL 1).
The ultrasonic fatigue test is a powerful tool which can realize an ultrafast fatigue test at 20 kHz 200 cycles higher than the normal, and performs a gigacycle fatigue test 109 cycles or more.
In a turbine blade, such as a gas turbine, it is known that kHz order vibration occurs during operation. In simulating fatigue due to fast vibration, it can be said that the ultrasonic fatigue test is an optimum testing method.
From this point, it is desirable to develop an ultrasonic fatigue testing device which can perform a test at high temperature. However, since this method is special, in realizing a test at high temperature, there are many problems to be solved.
The ultrasonic fatigue test is a testing method in which the entire system having a vibrator, a horn configured to expand amplitude, and a test specimen is in a resonance state to cause stress amplitude in the test specimen. FIG. 12 is a schematic view of the basic configuration of a high-temperature ultrasonic fatigue testing device. An ultrasonic fatigue testing device 100 includes horns 103 which are attached to both ends of a test specimen 106, an ultrasonic oscillation device 102 which is arranged in one horn 103, a load frame 128 which applies an average load, and a high frequency heating coil 127 which heats the test specimen 106.
In this case, the output of the ultrasonic oscillator is adjusted, and displacement amplitude (end surface displacement) of the end surface of the test specimen is controlled, thereby adjusting stress amplitude to be applied to the test specimen.
The relationship between end surface displacement and stress amplitude can be obtained by solving Expression (1) (NPL 1).
                    [                  Math          .                                          ⁢          1                ]                                                                                                U              ″                        ⁡                          (              x              )                                +                                    P              ⁡                              (                x                )                                      ⁢                                          U                ′                            ⁡                              (                x                )                                              +                                    k              2                        ⁢                          U              ⁡                              (                x                )                                                    =        0                            (        1        )                                [                  Math          .                                          ⁢          2                ]                                                                      P          ⁡                      (            x            )                          =                                            S              ′                        ⁡                          (              x              )                                            S            ⁡                          (              x              )                                                          (        2        )                                [                  Math          .                                          ⁢          3                ]                                                            k        =                  2          ⁢          π          ⁢                                          ⁢          f          ⁢                                    ρ              E                                                          (        3        )            
Here, U(x) denotes displacement amplitude, S(x) denotes a cross-sectional area, f denotes a resonance frequency, E denotes Young's modulus, and ρ denotes density.
When an hourglass test specimen shown in FIG. 13 is used, the cross-sectional area S(x) is expressed by Expressions (4) and (5).
                    [                  Math          .                                          ⁢          4                ]                                                                                  S            ⁡                          (              x              )                                =                                                    π                4                            ⁢              d              ⁢                                                          ⁢                              1                2                            ⁢                                                cosh                  2                                ⁡                                  (                                      α                    ⁢                                                                                  ⁢                    x                                    )                                            ⁢              α                        =                                          1                                  L                  ⁢                                                                          ⁢                  1                                            ⁢                                                cosh                                      -                    1                                                  ⁡                                  (                                                            d                      ⁢                                                                                          ⁢                      2                                                              d                      ⁢                                                                                          ⁢                      1                                                        )                                                                    ,                                  ⁢                  0          ≤          x          ≤                      L            ⁢                                                  ⁢            1                                              (        4        )                                [                  Math          .                                          ⁢          5                ]                                                                      S          ⁡                      (            x            )                          =                                            π              4                        ⁢            d            ⁢                                                  ⁢                          2              2                        ⁢            L            ⁢                                                  ⁢            1                    ≤          x          ≤                                    L              ⁢                                                          ⁢              1                        +                          L              ⁢                                                          ⁢              2                                                          (        5        )            
Here, d1 denotes a minimum part diameter, d3 denotes a shoulder part diameter, L1 denotes an R part half-length, and L2 denotes a shoulder part length.
When Expressions (1) to (5) are solved, the number of undetermined coefficients becomes four. Meanwhile, there are five environmental conditions to be satisfied including 1) displacement amplitude at the center of the test specimen is zero, 2) strain amplitude in the end surface of the test specimen is zero, 3) end surface displacement to which displacement amplitude is input at the same position, and continuity of 4) strain amplitude and 5) displacement amplitude in the boundary of an R part and a shoulder part.
That is, since the number of environmental conditions is greater than the number of undetermined coefficients by one, one of the dimensional parameters of the test specimen is restricted. This becomes a resonance condition in designing the test specimen, and the shoulder part length L2 which is most readily adjusted usually conforms to the resonance condition.